googologywikiaorg-20200223-history
Talk:Bachmann-Howard ordinal
BEAF visualization? FB100Z • talk • 17:02, April 4, 2013 (UTC) Since BEAF ends at \(\vartheta((\Omega^\Omega)\omega)\), BHO can't be visualised in that system. However, in Bird's array notation it can be represented as \(\{\omega,\omega [1 [1 \ldots [1 \backslash_{\omega} 1 \backslash_{\omega} 2 \ldots 2] 2] 2] 2\}\) (with \(\omega\) pairs of brackets). Ikosarakt1 (talk ^ 19:09, April 4, 2013 (UTC) So, bird array grows faster than BEAF? Jiawhien (talk) 10:43, May 30, 2013 (UTC) :Shortly, yes. Ikosarakt1 (talk ^ ) 10:52, May 30, 2013 (UTC) What does the ε function here mean? Is εΩ+1=ω^ω^...ω^(εΩ+1)(with ω ω's), which is equal to φ(1,Ω+1), or εΩ+1=εΩ^εΩ^...^εΩ=Ω^Ω^...^Ω(with ω Ω's), which is equal to θ1(1,0)? {hypcos} (talk) 15:24, July 18, 2013 (UTC) One variant of ε function, used for definition of BHO, uses equation εα+1 = αααα... (with ω α's) Ikosarakt1 (talk ^ ) 15:42, July 18, 2013 (UTC) εΩ+1=ω^ω^...ω^(Ω+1)=Ω^Ω^...^Ω (yes, I know, abuse of notation) LittlePeng9 (talk) 16:20, July 18, 2013 (UTC) Both expressions represent the same ordinal, the (Ω+1)st fixed point of a -> ω^a, or the first fixed point larger than Ω. To see this, let x be that fixed point. Obviously Ω+1 < ω^(Ω+1), so Ω+1 < x. Also, if a < x, then ω^a < ω^x = x. So Ω+1, ω^(Ω+1), ω^ω^(Ω+1), ... are all less than x, and therefore y = lim (Ω+1, ω^(Ω+1), ω^ω^(Ω+1), ...) <= x. On the other hand, ω^y = lim (ω^(Ω+1), ω^ω^(Ω+1), ω^ω^ω^(Ω+1) ...) = y, so y is a fixed point of a -> ω^a and therefore y = x. Similarly, if a,b < x then a^b < x. Proof: First, we will prove that if a,b < x, then a + b < x. First, x must be a limit ordinal, as if x = y + 1, then ω^x = ω^(y+1) = ω^y * ω = y * ω > y + 1 = x. Now, suppose max(a,b) = a. Then if a,b < x, then a+1 < x, and so x > ω^(a+1) = a * ω > a * 2 > a + b. Similarly for max(a,b) = b. Finally, having proven a,b < x => a + b < x, we have a^b <= (ω^ω^a)^(ω^b) = ω^ω^(a+b) < ω^ω^x = x. So, since Ω < x, we have Ω^Ω, Ω^Ω^Ω, Ω^Ω^Ω^Ω, ... < x. So z = lim (Ω, Ω^Ω, Ω^Ω^Ω, ...) <= x. On the other hand, ω^z = lim (ω^Ω, ω^Ω^Ω, ω^Ω^Ω, ...) <= lim (Ω^Ω, Ω^Ω^Ω, Ω^Ω^Ω^Ω, ...) = z <= ω^z. So z is a fixed point of a -> ω^a, and so z = x. Thus, εΩ+1 is both lim (ω^(Ω+1), ω^ω^(Ω+1), ω^ω^ω^(Ω+1) ...) and lim (Ω, Ω^Ω, Ω^Ω^Ω, ...). Deedlit11 (talk) 16:31, July 18, 2013 (UTC) I can show that sequence with ω's (later: A, other sequence is B) is upper bounded by BHO, but I'm not sure about lower bound. For upper bound, we can see that ω^(Ω+1) = ω^Ω*Ω = Ω*Ω = Ω^2 < Ω^Ω. So A1 is lower than B2. Then we get An = w^(An-1) and Bn = Ω^(Bn-1). By obvious reasons, A grows no faster than B. Ikosarakt1 (talk ^ ) 18:40, July 18, 2013 (UTC) :Your A2=ω^(Ω+1)=ω^Ω*ω=Ω*ω (not ω^Ω*Ω), A3=ω^(Ω*ω)=(ω^Ω)^ω=Ω^ω>B1, and A4=ω^Ω^ω=ω^(Ω*Ω*Ω*…*Ω)=(ω^Ω)^(Ω*Ω*…*Ω)=Ω^Ω^ω>B2, A5=ω^Ω^Ω^ω=Ω^Ω^Ω^ω>B3. :An is lower bounded by Bn-2. {hypcos} (talk) 02:37, July 19, 2013 (UTC) BEAF visualization With more recent discoveries regarding the strength of BEAF, has a visualization been found? FB100Z • talk • :I don't see how ordinals past \varepsilon_0 can be expressed in BEAF and \omega 's. X-structure arithmetic differs from \omega -arithmetic because, for example, X^^(X+1) = X^(X^^X) \neq X^^X while replacing X to \omega 's gives another result. Ikosarakt1 (talk ^ ) 16:25, June 2, 2014 (UTC) Equation What's wrong with the first ordinal statistifying the equation \(\vartheta(\Omega^\alpha) = \vartheta(\alpha)\)? Wythagoras (talk) 05:20, June 2, 2014 (UTC) :I don't think there is anything wrong. LittlePeng9 (talk) 05:28, June 2, 2014 (UTC) :If you say "the first ordinal satisfying the equation \(\vartheta(\Omega^\alpha) = \theta(\alpha)\)", it would mean that \(\alpha=\varepsilon_{\Omega+1}\). Therefore, BHO is equal to \(\vartheta(\varepsilon_{\Omega+1})\). King2218 (talk) 05:47, June 2, 2014 (UTC) epsilon sub Omega plus one Edit: Disregard all of this section since i am incapable of reading Consider the function PT(a,b) = power tower of a, b high Is it reasonable that for countable ordinals α=>ζ0, the set described by Union (from n=1 to infinity): PT(α,n) (lets call this set PT(α) for this talk section) call is similar to a fundamental sequence for e_{α+1} ? I would think so, since even if the members are different from the usual definition of phi fundamental sequences which is α+1, ω^(α+1), ω^ω^(α+1), ... it can be seen to converge to the same limit. (i mean, fundamental sequences for e_{α+1}, not α) even though PT(α)n> e_{α+1}n which doesn't matter Now my problem is, for uncountable ordinals, for example Ω, it does not seem the case that PT(Ω,n) converges to the same limit as the sequence for e{Ω+1} for example psi(Ω*ω or ω^(Ω+1)) = Z_ω while psi(Ω^Ω) is Γ_0 (using this definition of psi https://en.wikipedia.org/wiki/Ordinal_collapsing_function ) the implication is that the notation ε_{Ω+1} is not appropiate for "first fixed point of α -> Ω^α" EDIT: I just read deedlit's earlier comment on this page, please disregard all of this as i shamefully walk away slowly Chronolegends (talk) 17:17, February 25, 2017 (UTC)